Đáp án: a.$Q=\dfrac{2x}{x-1}$
b.$x=\dfrac2k+1, k\in Z, k<-2$ hoặc $k>0$
Giải thích các bước giải:
a.Ta có :
$Q=(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1})\cdot (x+\sqrt{x})$
$\to Q=(\dfrac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\dfrac{\sqrt{x}-2}{(\sqrt{x}+1)(\sqrt{x}-1)})\cdot \sqrt{x}(\sqrt{x}+1)$
$\to Q=(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1})\cdot \sqrt{x}$
$\to Q=\dfrac{(\sqrt{x}+2)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-1)}\cdot \sqrt{x}$
$\to Q=\dfrac{x+\sqrt{x}-2-(x-\sqrt{x}-2)}{(\sqrt{x}+1)(\sqrt{x}-1)}\cdot \sqrt{x}$
$\to Q=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{(\sqrt{x}+1)(\sqrt{x}-1)}\cdot \sqrt{x}$
$\to Q=\dfrac{2\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}\cdot \sqrt{x}$
$\to Q=\dfrac{2x}{x-1}$
b.Để Q nguyên
$\to \dfrac{2x}{x-1}\in Z$
$\to \dfrac{2x-2+2}{x-1}\in Z$
$\to \dfrac{2(x-1)+2}{x-1}\in Z$
$\to 2+\dfrac{2}{x-1}\in Z$
$\to \dfrac{2}{x-1}\in Z$
$\to \dfrac{2}{x-1}=k, k\in Z, k\ne 0$
$\to x-1=\dfrac2k$
$\to x=\dfrac2k+1$
Mà $x>0\to \dfrac2k+1>0\to \dfrac2k>-1\to k<-2$ hoặc $k>0$
Vậy $x=\dfrac2k+1, k\in Z, k<-2$ hoặc $k>0$
